We are given the equation: \[ \tan^{-1}(x) + \cot^{-1}(x) = \frac{2\pi}{2} \]
Step 1: Rearrange the equation: We can rearrange the equation as: \[ \left( \tan^{-1}(x) + \cot^{-1}(x) \right)^2 - 2 \tan^{-1}(x) \left( \frac{2\pi}{2} - \tan^{-1}(x) \right) = \frac{8}{\left( \frac{5\pi}{2} \right)} \]
Step 2: Simplify further: Simplifying the above equation gives: \[ 2 \left( \tan^{-1}(x) \right)^2 - 2 \left( \frac{2\pi}{2} \right) \tan^{-1}(x) - \frac{8}{\left( \frac{3\pi}{2} \right)} = 0 \]
Step 3: Simplify the terms: This equation can be rewritten as: \[ 2 \left( \tan^{-1}(x) \right)^2 - 2 \pi \tan^{-1}(x) - \frac{8}{\left( \frac{3\pi}{2} \right)} = 0 \]
Step 4: Deduce the solution: From the equation, we can deduce that: \[ \tan^{-1}(x) = \frac{-4\pi}{3} \quad \text{or} \quad \frac{4\pi}{3} \]
Final Step: Solve for \(x\): Therefore, the solutions for \(x\) are: \[ x = -1 \]
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)